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DEtools

  

regular_parts

  

Find regular parts of a linear ode

 

Calling Sequence

Parameters

Description

Examples

Calling Sequence

regular_parts(L, y, t, [x=x0])

Parameters

L

-

linear homogeneous differential equation

y

-

unknown function to search for

t

-

name used as parametrization variable

x0

-

(optional) a rational, an algebraic number or infinity

Description

• 

The regular_parts function computes the minimal generalized exponents of L at the point x0 and the corresponding regular parts. These are operators L_e which result from L by replacing y(x) by exp(int(e, x))*y(x). The Newton polygon of L_e at x_0 has a segment of slope 0 and 0 is a root of the indicial polynomial.

• 

The equation Ly=0 must be homogeneous and linear in y and its derivatives, and its coefficients must be rational functions in the variable x.

• 

x0 must be a rational or an algebraic number or the symbol infinity. If x0 is not passed as argument, x0 = 0 is assumed.

• 

The output is a set of solutions which are of the form exp(int(e, x))*y where e is a minimal generalized exponent and y is given as DESol object.

• 

The command with(DEtools,regular_parts) allows the use of the abbreviated form of this command.

Examples

withDEtools:

odeyx2x2ⅆⅆxyx6x3ⅆ2ⅆx2yx2x4ⅆ3ⅆx3yx+x6ⅆ4ⅆx4yx

odeyx2x2ⅆⅆxyx6x3ⅆ2ⅆx2yx2x4ⅆ3ⅆx3yx+x6ⅆ4ⅆx4yx

(1)

Then 0 is a singular point of this equation. Newton polygon is:

newton_polygonode,yx,u

13,12u,1,2+u

(2)

There are slopes > 0 so 0 is an irregular singular point.

rregular_partsode,yx,t

rxt=12t3,yt=ⅇ3ttDESol2027t4+259t34627t2+536t2081t5yt+227t6+2627t523t243t4t+227t3ⅆⅆtyt+481t713t6+16t513t329t4ⅆ2ⅆt2yt+281t8+127t7127t5ⅆ3ⅆt3yt+1324t9ⅆ4ⅆt4yt,yt,xt=t,yt=ⅇ2tt9DESol3024t3+1230t2+141tyt+2016t4+802t3+120t2+8tⅆⅆtyt+432t5+132t4+12t3ⅆ2ⅆt2yt+36t6+6t5ⅆ3ⅆt3yt+t7ⅆ4ⅆt4yt,yt

(3)

yields two transformed differential equations:

ode1op1,selecttype,rhsr12,DESol1

ode12027t4+259t34627t2+536t2081t5yt+227t6+2627t523t243t4t+227t3ⅆⅆtyt+481t713t6+16t513t329t4ⅆ2ⅆt2yt+281t8+127t7127t5ⅆ3ⅆt3yt+1324t9ⅆ4ⅆt4yt

(4)

ode2op1,selecttype,rhsr22,DESol1

ode23024t3+1230t2+141tyt+2016t4+802t3+120t2+8tⅆⅆtyt+432t5+132t4+12t3ⅆ2ⅆt2yt+36t6+6t5ⅆ3ⅆt3yt+t7ⅆ4ⅆt4yt

(5)

These operators have a Newton polygon with slope 0:

newton_polygonode1,yt,u

0,u,1,u29u27,3,12+u

(6)

newton_polygonode2,yt,u

0,u,1,u3+6u2+12u+8

(7)

This can help to find closed-form solutions:

ode1x1215x9+38x66x3yx+3x818x5+6x2ⅆⅆxyx+3x43xⅆⅆxⅆⅆxyx+ⅆⅆxⅆⅆxⅆⅆxyx

ode1x1215x9+38x66x3yx+3x818x5+6x2ⅆⅆxyx+3x43xⅆ2ⅆx2yx+ⅆ3ⅆx3yx

(8)

rregular_partsode,yx,t

rxt=t,yt=ⅇ13t3tDESolt3ⅆ3ⅆt3yt,yt

(9)

Since the general solution of the regular part is a+b*x+c*x^2 for some constants a,b and c, we obtain the general solution of the original equation by taking into account the exponential transformation:

simplifysubsyx=ⅇ13x3xa+bx+cx2,ode

6ⅇ13x3cx11+3x3ⅇ13x3xcx2+bx+ax116ⅇ13x3bx1032x2ⅇ13x3xcx2+bx+ax106ⅇ13x3ax9+6xⅇ13x3xcx2+bx+ax9+38ⅇ13x3cx8+38ⅇ13x3bx7+32x2ⅇ13x3xcx2+bx+ax7+38ⅇ13x3ax618xⅇ13x3xcx2+bx+ax615ⅇ13x3cx515ⅇ13x3bx415ⅇ13x3ax3+3xⅇ13x3xcx2+bx+ax3+ⅇ13x3cx2+ⅇ13x3bx+ⅇ13x3ax11

(10)

See Also

DEtools

DEtools/formal_sol